Problem: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}5x+4y &= 3 \\ 5x-y &= 8\end{align*}$
Explanation: Begin by moving the $x$ -term in the second equation to the right side of the equation. $-y = -5x+8$ Divide both sides by $-1$ to isolate $y$ $y = {5x - 8}$ Substitute this expression for $y$ in the first equation. $5x+4({5x - 8}) = 3$ $5x + 20x - 32 = 3$ Simplify by combining terms, then solve for $x$ $25x - 32 = 3$ $25x = 35$ $x = \dfrac{7}{5}$ Substitute $\dfrac{7}{5}$ for $x$ back into the top equation. $5( \dfrac{7}{5})+4y = 3$ $7+4y = 3$ $4y = -4$ $y = -1$ The solution is $\enspace x = \dfrac{7}{5}, \enspace y = -1$.